Applied Mathematical Sciences, Vol. 8, 2014, no. 102, 5079 - 5082
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2014.4154
Inequalities among Related Triplets of Fibonacci Numbers
Sanjay Harne1, Bijendra Singh2, Gurbeer Kaur Khanuja2, Manjeet Singh Teeth3
1
2
Government Holkar Science College, Indore, M.P., India
School of Studies in Mathematics,Vikram University, Ujjain, M.P., India
3
M.B. Khalsa College, Indore, M.P., India
Copyright © 2014 Sanjay Harne et al. This is an open access article distributed under the Creative Commons
Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the
original work is properly cited.
Abstract
In this paper we consider Fibonacci inequalities and relate them through the sequence
m 
n
r r 1
defined by where n is a fixed natural number and F1 , F2 , F3 , ... are the ordinary
Fibonacci numbers.
Mathematics Subject Classification: 11B39
Keywords: Fibonacci sequence
INTRODUCTION
“Fibonacci inequalities” have been studied in a variety of contexts. Atanassov [3]
considered the Fibonacci inequalities and relate them through the sequence mr r 0 defined by
n
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Sanjay Harne et al.
mr  Fr Fn1r , where n is a fixed natural number and F1 , F2 , F3 , ....... are the ordinary Fibonacci
numbers as defined in [4]. We consider the similar Fibonacci inequalities and relate them
through the sequence mr r 1 defined by mr  Fr Fn1r Fn2r , where n is a fixed natural number
n
and F1 , F2 , F3 , ....... are the ordinary Fibonacci numbers.
Theorem: For every natural number k, the following inequalities for the elements of the
sequence mk k 1 are valid :
n
(a) For n = 4k,
(i) F1 F4k F4k 1  F3 F4k 2 F4k 1  ...  F2k 1 F2k F2k 1  F2k 2 F2k 1 F2k  F2k 4 F2k 3 F2k 2  ...  F4k F1 F2
(ii)
F2 F4k 1 F4k  F4 F4k 3 F4k 2  ..  F2k F2k 1 F2k 2  F2k 1 F2k F2k 1  F2k 3 F2k 2 F2k 1  ..  F4k 1 F2 F3
(b) For n=4k+1,
(i)
F1 F4k 1 F4k 2  F3 F4k 1 F4k  ..  F2k 1 F2k 1 F2k 2  F2k 2 F2k F2k 1  F2k 4 F2k 2 F2k 1  ..  F4k F2 F3
(ii)
F2 F4k F4k 1  F4 F4k 2 F4k 1  ..  F2k F2k 2 F2k 3  F2k 1 F2k 1 F2k 2  F2k 3 F2k 1 F2k  ..  F4k 1 F1 F2
(c) For n = 4k+2,
(i)
F1 F4k 2 F4k 3  F3 F4k F4k 1  ....  F2k 1 F2k 2 F2k 3  F2k 2 F2k 1 F2k 2  F2k 4 F2k 1 F2k  ....  F4k 2 F1 F2
(ii)
F2 F4k 1 F4k 2  F4 F4k 1 F4k  ....  F2k F2k 3 F2k 4  F2k 1 F2k 2 F2k 3  F2k 3 F2k F2k 1  ....  F4k 1 F2 F3
(d) For n = 4k+3,
(i)
F1 F4k 3 F4k 4  F3 F4k 1 F4k 2  ....  F2k 1 F2k 3 F2k 4  F2k 2 F2k 2 F2k 3  F2k 4 F2k F2k 1  ....  F4k 2 F2 F3
(ii)
F2 F4k 2 F4k 3  F4 F4k F4k 1  ....  F2k F2k 4 F2k 5  F2k 1 F2k 3 F2k 4  F2k 3 F2k 1 F2k 2  ....  F4k 3 F1 F2
Proof : case (a)
(i) F1 F4k 4 F4k 5  F3 F4k 2 F4k 3
 F4k 4 F4k 3  F42k 4  2F4k 2 F4k 3
Inequalities among related triplets of Fibonacci numbers
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 F42k 3  F42k 4  F4k 2 F4k 3
 F42k 4  F4k 3 ( F4k 3  F4k 2 )  0
Thus, F1 F4k 4 F4k 5  F3 F4k 2 F4k 3 .
This shows that the inequality (1.1) is valid for i = 1.
F2i 1 F4k 2i 6 F4k 2i 7  F2i 1 F4k 2i4 F4k 2i 5
..........(1.1)
Let us assume that, for some i, 1  i  k , the inequality (1.1) is true.
For desired result, we use induction method and prove that the inequality
F2i1 F4k 2i4 F4k 2i5  F2i3 F4k 2i2 F4k 2i3
..........(1.2)
is also true.
But, F2i 1 F4k 2i 4 F4k 2i 5  F2i 3 F4k 2i 2 F4k 2i 3
 2F42k 2i 3 F2i 1  3F4k 2i3 F4k 2i 2 F2i1  F42k 2i2 F2i 1  F2i2 F4k 2i2 F4k 2i3  F2i 1 F4k 2i2 F4k 2i 3
 2F42k 2i 3 F2i1  F42k 2i 2 F2i1  (2F2i 1  F2i 2 ) F4k 2i 2 F4k 2i 3
 2F42k 2i 3 F2i 1  F42k 2i 2 F2i 1  ( F2i 1  F2i ) F4k 2i 2 F4k 2i 3  0
Now, (ii ) F2 F4k 3 F4k 4  F4 F4k 1 F4k 2
 ( F4k 1  F4k 2 ) ( F4k 1  2 F4k 2 )  3 F4k 1 F4k 2
 F42k 1  2 F42k 2  0
Thus, F2 F4k 3 F4k 4  F4 F4k 1 F4k 2
This shows that the inequality (1.3) is valid for i = 1.
F2i F4k 2i5 F4k 2i6  F2i 2 F4k 2i3 F4k 2i4
..........(1.3)
Let us assume that for some i , 1  i  k the inequality (1.3) is true.
Then we must prove that the inequality (1.4) is also true.
F2i 2 F4k 2i3 F4k 2i4  F2i4 F4k 2i 1 F4k 2i 2
..........(1.4)
But, F2i 2 F4k 2i 3 F4k 2i 4  F2i 4 F4k 2i 1 F4k 2i 2
 F2i 2 ( F4k 2i1  F4k 2i 2 ) (2 F4k 2i2  F4k 2i1 )  ( F2i1  2 F2i2 ) F4k 2i1 F4k 2i2
 F2i  2 (3 F4k 2i 1 F4k 2i  2  F42k 2i 1  2 F42k 2i  2 )  ( F2i 1  2 F2i  2 ) F4k 2i 1 F4k 2i  2  0
This shows that the inequality (1.4) is true and hence (ii) is true.
Particularly, when k = 3, n becomes 12 which gives the following two results:
1) F1 F12 F13  F3 F10 F11  F5 F8 F9  F7 F6 F7  F8 F5 F6  F10 F3 F4  F12 F1 F2
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2) F2 F11F12  F4 F9 F10  F6 F7 F8  F7 F6 F7  F9 F4 F5  F11F2 F3
Similarly other cases can be proved.
Corollary : For every natural number n the maximal element[1] and [2] of the sequence
mk nk1 is
F1 Fn Fn1 and the minimal element is Fn F1 F2 .
REFERENCES
1. A.F. Alameddine, “Bounds on the Fibonacci Number of a Maximal Outerplanar Graph”, The
Fibonacci Quarterly 36.3(1998): 206-10.
2. K.T. Atanassov. “One Extremal Problem” Bulletin of Number Theory & Related Topics 8.3
(1984): 6-12.
3. K.T. Atanassov, Ron Knott, Kiyota Ozeki, A.G. Shannon, Laszlo Szalay, “Inequalities among
related pairs of Fibonacci Numbers”, The Fibonacci Quarterly (Feb. 2003):20-22.
4. V.E. Hoggatt,Jr. Fibonacci and Lucas Numbers, p. 59, Boston: Houghton-Mifflin, 1969.
Received: January 15, 2014

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